MATHHX B
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9.4 Lineær programmering uden computer
Vi har gjort brug af GeoGebra til at regne opgaverne, men det er nemt at regne dem uden GeoGebra. Det tager bare lidt længere tid.
Vi vender tilbage til det første eksempel i første afsnit og regner det nu uden computer:
En tøjproducent producerer trøjer og bukser.
Producenten har \(\SI {120}{m^2}\) bomuld og \(\SI {130}{m^2}\) polyester på lager.
Der går \(\SI {1,5}{m^2}\) bomuld og \(\SI {0,5}{m^2}\) polyester til en trøje.
Der går \(\SI {1}{m^2}\) bomuld og \(\SI {2}{m^2}\) polyester til et par et par bukser.
Dækningsbidraget er \(300\) kr. for en trøje og \(400\) kr. for et par bukser.
Hvor mange trøjer og hvor mange bukser skal der produceres få det størst mulige samlede dækningsbidrag? Hvad bliver det maksimale dækningsbidrag?
Opstilling af skema
. |
|
Trøjer |
Bukser |
Til rådighed |
Bomuld |
\(\SI {1,5}{m^2}\) |
\(\SI {1}{m^2}\) |
\(\SI {120}{m^2}\) |
Polyester |
\(\SI {0,5}{m^2}\) |
\(\SI {2}{m^2}\) |
\(\SI {130}{m^2}\) |
Dækningsbidrag |
\(300\) kr. |
\(400\) kr. |
|
|
|
|
|
Kriteriefunktionen
Vi sætter
\(x=\) Antal producerede trøjer
\(y=\) Antal producerede bukser
Dækningsbidraget må dermed være givet ved:
\[f(x,y)=300x+400y\]
Polygonområde
Ud fra skemaet ses at
\[1{,}5x+y\leq 120\]
For polyester må der gælde:
\[0{,}5x+2y\leq 130\]
Da vi ikke kan producere et negativt antal trøjer eller bukser har vi:
\[x\geq 0 \quad \text {og}\quad y\geq 0\]
Vi skal tegne polygonområdet, men denne gang med papir og blyant. Vi starter med den første ulighed og isolerer \(y\):
\(\seteqnumber{0}{9.}{0}\)
\begin{align*}
1{,}5x+y & \leq 120\\ y & \leq -1{,}5x+ 120
\end{align*}
Vi kan nu tegne uligheden. I første omgang lader vi som om, der står et lighedstegn i stedet for ulighedstegnet. Vi kan så tegne linjen ud fra hældning (\(-1{,}5\)) og skæringspunkt med y-aksen (\(120\)):
Vi isolerer \(y\) i den næste ulighed, hvilket giver \(y\leq -0{,}25x+65\), og tegner den ind:
Nu husker vi at der stod \(y\leq \ldots \) i begge de to uligheder. Det betyder at vi er begræset til de punkter som ligger under linjerne (de må godt ligge på en linjen, men ikke over en linje). Vi husker også ulighederne \(x\geq
0\) og \(y\geq 0\), hvilket betyder at vi er begrænset til punkter, som har ikke-negative koordinater. Det giver os et område vi kan farve blåt:
Niveaulinjer
Vi skal nu tegne niveaulinjer. Det er den mest problematiske del fordi det kræver, at man prøver sig lidt frem før man finder nogle gode linjer.
Vi vil nu tegne \(N(20000)\) og \(N(30000)\). For at finde dem skal vi sætte \(f(x,y)=20000\) og \(f(x,y)=30000\). Det giver:
\[300x+400y=20000\quad \text {og} \quad 300x+400y=30000\]
Isolerer vi \(y\) i disse ligninger fås
\[y=-0{,}75x + 50\quad \text {og} \quad y= -0{,}75x + 75\]
Disse to linjer kan vi nu tegne ind:
Ud fra niveaulinjerne ses det at niveauerne vokser skråt op til højre, og at det optimale punkt således ligger i skæringspunktet mellem de to røde begrænsningslinjer.
Bestemmelse af det optimale punkt
Nu skal vi bare bestemme koordinaterne til det punkt vi har fundet. Vi har allerede isoleret \(y\) i de to røde begrænsningslinjer og vi fik (når man erstatter ulighedstegn med lighedstegn):
\[y=-1{,}5x+ 120 \quad \text {og} \quad y = -0{,}25x+65\]
Vi regner nu skæringspunktet ved at sætte de to funktionsudtryk lig hinanden.
\[-1{,}5x+ 120 = -0{,}25x+65\]
Løser vi den ligning får vi \(x=44\). Vi kan nu regne \(y\)-værdien ved at sætte \(44\) ind i en af de to forskrifter for begrænsningslinjerne:
\[y=-1{,}5\cdot 44 + 120=54\]
Skæringspunktet er altså \((44,54)\). Det betyder at vi skal producerer \(44\) trøjer og \(54\) bukser for at få det maksimale dækningsbidrag. Det maksimale dækningsbidrag finder vi ved at sætte punktets koordinater ind i \(f\):
\[f(44,54)=300\cdot 44+400\cdot 54=34800\]
Vi konkluderer, at det optimale dækningsbidrag er \(34800\) kr.
Øvelse 9.4.1
En elev vil sælge slikposer i kantinen. Hun reklamerer med:
Billige slikposer. Kom og køb. Du får miniskumbananer og/eller piratos. Mindst 20 stykker slik i hver pose. Mindst 120 g i posen!
En miniskumbanan vejer \(9\) gram og en piratos vejer \(4\) gram.
Eleven kan indkøbe miniskumbananer til \(2\) kr. pr. stk. og piratos til \(50\) øre pr. stk.
Eleven vil gerne minimere sine omkostninger.
-
a) Bestem (uden GeoGebra), hvad eleven skal putte i poserne og bestem den samlede udgift pr. pose.
-
b) Pludselig falder prisen på miniskumbananer til \(0{,}75\) kr. stk. Hvad skal hun nu putte i poserne og hvad vil det koste hende? GeoGebra er stadig
haram.
-
c) Prisen på miniskumbananer falder nu yderligere til \(0{,}25\) kr. stk. Hvad skal hun nu putte i poserne og hvad vil det koste hende? GeoGebra er
stadig haram.
Løsning 9.4.1
-
a) Hun skal putte \(30\) piratos i hver pose og det vil kosten hende \(15\) kr. pr. pose.
-
b) Hun skal putte \(8\) miniskumbananer og \(12\) piratos i hver pose og det vil koste hende \(12\) kr. pr. pose.
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c) Hun skal putte \(20\) miniskumbananer i hver pose og det vil koste hende \(5\) kr. pr. pose.